An arrow is shot at an angle of above the horizontal. The arrow hits a tree at an angle 45 degrees horizontal distance D = 220 away, at the same height above the ground as it was shot. Use g = 9.8 for the magnitude of the acceleration due to gravity.
Find t_a, the time that the arrow spends in the air.
Suppose someone drops an apple from a vertical distance of 6.0 meters, directly above the point where the arrow hits the tree.
How long after the arrow was shot should the apple be dropped, in order for the arrow to pierce the apple as the arrow hits the tree?
Please provide formulas and numeric answers
Thank You!
at 45degrees, the final vertical velocity is the same as the horizontal velocity. So the same applies to launch.
Horizontal analysis:
distance=Vhorizontal*timeinair.
220=Vhorizontal*time
time=220/Vh
Vertical analysis:
hf=hi+Vv*t-1/2 g t^2
0=0+Vv*220/Vh-4.9 (220/Vh)^2
because Vv=Vh
220=4.9*220^2/Vh^2
solve for Vh
then, solve for timeinair from the first equation.
How long does it take for the apple to fall? 6=gt or t=6/g
subtract that from the time in air above, and you have the time after launch for the apple to be released.
To find the time that the arrow spends in the air (t_a), we can use the equation of motion in the vertical direction:
y = v_y0 * t - (1/2) * g * t^2
Since the arrow is shot at the same height from where it hits the tree, the initial vertical velocity (v_y0) will be zero. Also, at the highest point of its trajectory, the arrow will be at 45 degrees with the horizontal. At this point, the vertical and horizontal components of the velocity will have equal magnitudes, as they are at right angles. Therefore, we can write:
v_y0 = v0 * sin(45)
v_x0 = v0 * cos(45)
Where v0 is the initial velocity of the arrow.
At the time the arrow hits the tree, the height (y) will be zero. Therefore, we can write:
0 = v_y0 * t_a - (1/2) * g * t_a^2
Simplifying this equation, we have:
(1/2) * g * t_a^2 = v_y0 * t_a
(1/2) * g * t_a = v0 * sin(45) * t_a
Solving for t_a, we get:
t_a = (v0 * sin(45)) / ((1/2) * g)
Given that the horizontal distance D = 220, we can find the initial velocity using the horizontal component of the velocity:
D = v_x0 * t_a
220 = v0 * cos(45) * t_a
Simplifying this equation, we have:
v0 * cos(45) = 220 / t_a
v0 = (220 / t_a) / cos(45)
Now, let's consider the apple being dropped from a vertical distance of 6.0 meters above the point where the arrow hits the tree. The apple will fall vertically, so we can use the equation of motion:
y_apple = (1/2) * g * t_d^2
Where y_apple = 6.0 meters and t_d is the time after the arrow was shot until the apple is dropped.
At the time the apple hits the ground, the height will also be zero. Therefore, we can write:
0 = (1/2) * g * t_d^2
Simplifying this equation, we have:
t_d = sqrt((2 * y_apple) / g)
Therefore, in order for the arrow to pierce the apple as it hits the tree, the apple should be dropped t_d seconds after the arrow was shot.
To calculate numeric values, we need to know the initial velocity of the arrow.
To solve this problem, we can use the basic principles of projectile motion.
First, let's find t_a, the time that the arrow spends in the air.
We know that the range of the arrow, R, is given by:
R = (v₀² * sin(2θ)) / g
where v₀ is the initial velocity of the arrow and θ is the launch angle.
We are given the horizontal distance, D, which is equal to the range of the arrow:
D = R
So, we can rearrange the formula to solve for v₀:
v₀ = √((D * g) / sin(2θ))
Substituting the given values:
v₀ = √((220 * 9.8) / sin(90)) [Since sin(2θ) = sin(90º) = 1]
v₀ = √(2156)
v₀ ≈ 46.45 m/s
Now, we can use the equation for projectile motion to find the time, t_a, it takes for the arrow to reach the tree:
D = v₀ * cos(θ) * t_a
Rearranging the formula, we can solve for t_a:
t_a = D / (v₀ * cos(θ))
Substituting the given values:
t_a = 220 / (46.45 * cos(45º)) [Since cos(θ) = cos(45º) = 1/sqrt(2)]
t_a ≈ 4.15 seconds
Now, let's move on to the second part of the problem.
To find how long after the arrow was shot the apple should be dropped, we need to consider the relative vertical motion of the apple and the arrow.
The time it takes for the arrow to reach the tree is t_a = 4.15 seconds.
During this time, the apple will fall vertically from a height of 6.0 meters.
Using the equation for vertical displacement, s = (1/2) * g * t², we can solve for the time it takes for the apple to fall:
6.0 = (1/2) * 9.8 * t²
Rearranging the formula and solving for t:
t = √(12 / 9.8)
t ≈ 1.22 seconds
Therefore, the apple should be dropped approximately 1.22 seconds after the arrow is shot for the arrow to pierce the apple as the arrow hits the tree.